Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $t = \dfrac{k - 1}{k^2 + 6k - 7} \times \dfrac{k^2 - 2k - 63}{k - 7} $
Answer: First factor out any common factors. $t = \dfrac{k - 1}{k^2 + 6k - 7} \times \dfrac{k^2 - 2k - 63}{k - 7} $ Then factor the quadratic expressions. $t = \dfrac {k - 1} {(k + 7)(k - 1)} \times \dfrac {(k + 7)(k - 9)} {k - 7} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac {(k - 1) \times (k + 7)(k - 9) } { (k + 7)(k - 1) \times (k - 7)} $ $t = \dfrac {(k + 7)(k - 9)(k - 1)} {(k + 7)(k - 1)(k - 7)} $ Notice that $(k + 7)$ and $(k - 1)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {\cancel{(k + 7)}(k - 9)(k - 1)} {\cancel{(k + 7)}(k - 1)(k - 7)} $ We are dividing by $k + 7$ , so $k + 7 \neq 0$ Therefore, $k \neq -7$ $t = \dfrac {\cancel{(k + 7)}(k - 9)\cancel{(k - 1)}} {\cancel{(k + 7)}\cancel{(k - 1)}(k - 7)} $ We are dividing by $k - 1$ , so $k - 1 \neq 0$ Therefore, $k \neq 1$ $t = \dfrac {k - 9} {k - 7} $ $ t = \dfrac{k - 9}{k - 7}; k \neq -7; k \neq 1 $